What is the d/(dy)(1+ye^{xy}) ?

The d/(dy)(1+ye^{xy}) is e^{yx}+e^{yx}yx

What is the first d/(dy)(1+ye^{xy}) ?

The first d/(dy)(1+ye^{xy}) is e^{yx}+e^{yx}yx

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\frac{d}{d y} (1 + y e^{x y})

e^{y x} + e^{y x} y x

Solution steps

\frac{d}{d y} (1 + y e^{x y})

Treat

x

as a constant

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d y} (1) + \frac{d}{d y} (y e^{x y})

\frac{d}{d y} (1) = 0

\frac{d}{d y} (y e^{x y}) = e^{x y} + x e^{x y} y

= 0 + e^{x y} + x e^{x y} y

Simplify

= e^{y x} + e^{y x} y x

\int x tan (π x) d x

\frac{\partial}{\partial x} (sin (4 x - 2 y))

\frac{d}{d x} (\frac{x}{e ^{x} + 1})

\frac{\partial}{\partial x} (x e^{- 3 x y})

n = 1 \sum \infty \frac{n}{3}

What is the d/(dy)(1+ye^{xy}) ?
The d/(dy)(1+ye^{xy}) is e^{yx}+e^{yx}yx
What is the first d/(dy)(1+ye^{xy}) ?
The first d/(dy)(1+ye^{xy}) is e^{yx}+e^{yx}yx